Question

A 200 gram of ice block falls from a height of 80 meters onto a bucket of 1 kg water in a copper bucket of mass 100 grams. The initial temperature of the ice is 0 and the bucket and water is 20 C. Assume that 60% of the potential energy of the ice block goes into heat!!!! What is the final temperature of the ice, water and bucket mixture?? If not all the ice melts, then what is the final mass of the remaining block of ice?

Use C_cu = .092 cal/g.C, C_water = 1 cal/g.C, L_fwater = 333.55 J/gram

answer choices

a) 4.02 C

b) 61.6 grams of ice block left

c) 45.5 grams of ice block left

d) 3.53 C

e) 3.50 C

Answer 1

d) 3.53 C

let
m_Ice = 200 grams
m_copper = 100 grams
m_water = 1 kg

Heat supplied to the system due to loss of potentuao energy, Q_supplied = 0.6*m_Ice*g*h
= 0.6*0.2*9.8*80

= 94.08 J

Heat required to melt the total ice, Q1 = 200*Lf_water

= 200*333.55

= 66710 J

Heat lost by water and copper when it comes from 20 C to 0 C, Q_lost = m_Copper*C_copper*(20 - 0) + m_water*C_water*(20 - 0)

= 100*0.092*4.186*(20 - 0) + 1000*1*4.186*(20 - 0)

= 84490 J

clearly, 66710 J < 84490 J

so, total Ice melts

let T is the final temperature.

Heat gained by ice = heat supplied due to loss of potential energy + heat lost by water and copper

m_Ice*Lf_water + m_Ice*C_water*(T - 0 ) = 94.08 + m_Copper*C_copper*(20 - T) + m_water*C_water*(20 - T)

200*333.55 + 200*4.186*(T - 0) = 94.08 + 100*0.092*4.186*(20 - T) + 1000*1*4.186*(20 - T)

on solving the above equation we get

T = 3.53 C <<<<<<<<----------------Answer